All categories

2 years ago

There are 8 pencils in package.how many packages will be needed for 28 children if each child get 4 pencils

1 answer:

7 0

8 pencils = 1 package
28 children need pencils
Each child gets 4 pencils

28x4 = 112 (pencils needed)

112/8 = 14 packages needed.

You might be interested in

Simplify the expression.<br> 6y + 4/15 + 9m + 1/15 + 7y - 3m =

coldgirl [10]

Answer:

13y + 1/3 + 6m

Step-by-step explanation:

6y + 4/15 + 9m + 1/15 + 7y - 3m = 6y +1/3 + 9m + 7y - 3m = 13y + 1/3 + 6m

4 0

2 years ago

Read 2 more answers

The Bobcat baseball team has a total of 53 games this season. There are 9 games in the first month. After the first month, the t

lisabon 2012 [21]

Answer:

The season will last 4 months after the first month (5 months in total)

Step-by-step explanation:

53 games

9 games first month

11 games per month (since the second one)

x - number of months

9 + 11x = 53

11x = 53-9

11x = 44

x = 4

5 0

2 years ago

I need help with the answers

vlada-n [284]

Answer:

i dont get it

Step-by-step explanation:

4 0

2 years ago

HELP ME I DON'T UNDERSTAND THIS :(

andriy [413]

Answer:

Step-by-step explanation:

Since the figures are similar then the ratios of corresponding sides are equal, that is

$\frac{AD}{WZ}$ = $\frac{CD}{YZ}$ , substitute values

$\frac{10}{46}$ = $\frac{5}{YZ}$ ( cross- multiply )

10YZ = 230 ( divide both sides by 10 )

YZ = 23 → c

3 0

3 years ago

Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent

goblinko [34]

Answer:

$y(x)=x^2+5x$

Step-by-step explanation:

Given: $y'=\sqrt{4y+25}$

Initial value: y(1)=6

Let $y'=\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\sqrt{4y+25}$

Variable separable

$\dfrac{dy}{\sqrt{4y+25}}=dx$

Integrate both sides

$\int \dfrac{dy}{\sqrt{4y+25}}=\int dx$

$\sqrt{4y+25}=2x+C$

Initial condition, y(1)=6

$\sqrt{4\cdot 6+25}=2\cdot 1+C$

$C=5$

Put C into equation

Solution:

$\sqrt{4y+25}=2x+5$

$4y+25=(2x+5)^2$

$y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$

$y(x)=x^2+5x$

Hence, The solution is $y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}$ or $y(x)=x^2+5x$

4 0

2 years ago